Acceleration of Gravity problem?

Question

A certain freely falling object requires 1.50 seconds to travel to the last 30 meters before it hits the ground. From what height above the ground does it fall?

Answer 1

No matter what height you drop and object from, the speed is constant.

Answer 2

38.16 m

Answer 3

I get 38.15925 meters.

Answer 4

Formula is

S = VT + (1/2)gT^2

where

S = 30 meters (given)

V = velocity of the object as it starts its last 30 meters of fall

T = time = 1.5 sec.

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting values,

30 = (1.5)V + (1/2)(9.8)(1.5^2)

30 = 1.5V + 11.025

1.5V = 18.975

V = 12.65 m/sec. --- this is the velocity of the object as it  starts its descent for the last 30 meters.

The next equation to use is

V^2 - Vo^2 = 2gs

where

V = 12.65 m/sec (as calculated above)

Vo = initial velocity = 0 (assuming it was simply dropped)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

s = height that object travels before the last 30 meters

Substituting values,

12.65^2 - 0 = 2(9.8)(s)

Solving for "s",

s = 12.65^2/(2*9.8)

s = 8.16 meters

Therefore, initial height from which object was released = 30 + 8.16 = 38.16 meters  

Answer 5

45 meters  

Answer 6

We will split up the fall in 2 sections the last 30m (s) of which we know & the first section(s`) of which we are yet to know.

s = ut + (1/2) gt^2

30 = 1.5u + (9.8/2)(1.5)^2

30 = 3/2 u + 4.9(9/4)

u = (120 - 44.1)/6 = 12.65 m/s

This is the velocity at the start of the last 30m fall.

Now,

2gs` = u^2 - v^2;  v = 0(initial velocity)

19.6 s` = 160.0225

s` = 160.0225/19.6

= 8.1644m

Total height of fall = s + s`

= 30 + 8.1644 m

= 38.1644 m